Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Cross Product, Area and Torque

The cross product takes two vectors in 3D and gives back a third vector, perpendicular to both. Its length is an area, which is why it shows up for parallelograms, triangles and torque.

How It Works

The formula. For $\vec{u} = [u_1, u_2, u_3]$ and $\vec{v} = [v_1, v_2, v_3]$ :

$\vec{u} \times \vec{v} = [u_2v_3 - u_3v_2, ; u_3v_1 - u_1v_3, ; u_1v_2 - u_2v_1]$ .

To remember it, write $\vec{u}$ above $\vec{v}$ with each set of components written twice, and cross out the first and last columns. For each neighbouring pair of columns left, multiply top-left by bottom-right and subtract top-right times bottom-left. Always check your answer: its dot product with each original vector must be zero.

Direction. The right-hand rule fixes which way it points: fingers along the first vector, curl them toward the second, and your thumb points along the cross product. That is why order matters. Swapping the vectors flips the direction, so $\vec{v} \times \vec{u} = -(\vec{u} \times \vec{v})$ . A vector crossed with itself, or with any parallel vector, gives the zero vector.

Length is area. $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta$ . Picture the parallelogram with sides $\vec{u}$ and $\vec{v}$ : its base is $|\vec{u}|$ and its height is $|\vec{v}|\sin\theta$ , so the cross product's length is exactly that parallelogram's area. A triangle is half a parallelogram. For a triangle with vertices $A$ , $B$ and $C$ , build two side vectors from the same corner and take half the length of their cross product.

Torque is the turning effect of a force. With $\vec{r}$ running from the pivot to where the force $\vec{F}$ is applied, $\vec{\tau} = \vec{r} \times \vec{F}$ and its magnitude is $|\vec{r}||\vec{F}|\sin\theta$ , measured in newton metres. The angle is between the lever arm and the force, tail to tail. Pushing at $90^\circ$ gives the most torque; pushing along the arm gives none.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find $\vec{u} \times \vec{v}$ for $\vec{u} = [1, 2, 3]$ and $\vec{v} = [4, 0, -1]$ , and check it.
Show solution
1. First component: $2(-1) - 3(0) = -2$ . Second: $3(4) - 1(-1) = 13$ . Third: $1(0) - 2(4) = -8$ .
2. Check with $\vec{u}$ : $-2 + 26 - 24 = 0$ . Check with $\vec{v}$ : $-8 + 0 + 8 = 0$ .
Answer: $\vec{u} \times \vec{v} = [-2, 13, -8]$ , perpendicular to both.
Easy
$|\vec{u}| = 5$ , $|\vec{v}| = 8$ and the angle between them is $30^\circ$ . Find $|\vec{u} \times \vec{v}|$ .
Show solution
1. $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}|\sin\theta = (5)(8)\sin 30^\circ = 40\left(\frac{1}{2}\right) = 20$ .
Answer: $20$
Medium
Find the area of the parallelogram with sides $\vec{a} = [2, 1, 0]$ and $\vec{b} = [1, 3, 2]$ .
Show solution
1. $\vec{a} \times \vec{b} = [1(2) - 0(3), ; 0(1) - 2(2), ; 2(3) - 1(1)] = [2, -4, 5]$ .
2. Area $= |\vec{a} \times \vec{b}| = \sqrt{4 + 16 + 25} = \sqrt{45} = 3\sqrt{5}$ .
Answer: $3\sqrt{5} \approx 6.71$ square units
Medium
Find the area of the triangle with vertices $A(1, 0, 2)$ , $B(3, 1, 2)$ and $C(0, 2, 5)$ .
Show solution
1. Side vectors from $A$ : $\vec{AB} = [2, 1, 0]$ and $\vec{AC} = [-1, 2, 3]$ .
2. $\vec{AB} \times \vec{AC} = [1(3) - 0(2), ; 0(-1) - 2(3), ; 2(2) - 1(-1)] = [3, -6, 5]$ .
3. $|\vec{AB} \times \vec{AC}| = \sqrt{9 + 36 + 25} = \sqrt{70}$ , the parallelogram's area.
4. The triangle is half: $\frac{\sqrt{70}}{2}$ .
Answer: $\frac{\sqrt{70}}{2} \approx 4.18$ square units
Test level
A $30$ cm wrench is pushed at its end with a force of $60$ N at $75^\circ$ to the handle. Find the magnitude of the torque. What is the smallest force that would give $20$ N·m with this wrench?
Show solution
1. Convert to metres: $|\vec{r}| = 0.3$ m.
2. $|\vec{\tau}| = |\vec{r}||\vec{F}|\sin\theta = (0.3)(60)\sin 75^\circ \approx 18(0.9659) \approx 17.4$ N·m.
3. The smallest force is the one applied at $90^\circ$ , where $\sin\theta = 1$ : $20 = 0.3|\vec{F}|$ , so $|\vec{F}| = \frac{20}{0.3} \approx 66.7$ N.
Answer: About $17.4$ N·m; at least $66.7$ N, pushed at $90^\circ$ to the handle.

Learn It on Martian Lab

Martian Lab teaches this in the Cross Product lesson of its MCV4U Cartesian Vectors unit. Watch it, then take the Cartesian Vectors practice tests to see if it stuck.

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More Cartesian Vectors topics: Dot Product and Projections

More MCV4U topics: Evaluating Limits · Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Equations of Planes · Intersection of Three Planes

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